Directional Derivatives of Functions from Rn To Rm

Directional Derivatives of Functions from Rn To Rm

We recent defined what the concept of a partial derivative of a function $\mathbf{f} : S \to \mathbb{R}^m$ (where $S \subseteq \mathbb{R}^n$ is open). We now extend this definition to the more general directional derivative.

 Definition: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$. Let $\mathbf{u} \in \mathbb{R}^n$. Then the Directional Derivative of $\mathbf{f}$ at $\mathbf{c}$ in the Direction of $\mathbf{u}$ is defined as $\displaystyle{\mathbf{f}'(\mathbf{c}, \mathbf{u}) = \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h \mathbf{u}) - \mathbf{f}(\mathbf{c})}{h}}$ provided that this limit exists.

There are a few important things to point out. First, $\mathbf{f}$ is a function which maps elements (vectors) in an open subset $S$ of $\mathbb{R}^n$ to elements (vectors) in $\mathbb{R}^m$. Secondly, if $\mathbf{u} = 0$ then $\mathbf{f}'(\mathbf{c}, \mathbf{u}) = \mathbf{f} (\mathbf{c}, \mathbf{0}) = \mathbf{0}$. Thirdly, note that $\mathbf{f}'(\mathbf{c}, \mathbf{u})$ is itself a vector in $\mathbb{R}^m$!

If $\mathbf{u} = \mathbf{e}_k$ then the definition of the directional derivative of $\mathbf{f}$ in the direction of the unit vector in the direction of the $k^{\mathrm{th}}$ coordinate axis is simply the partial derivative of $\mathbf{f}$ with respect to the $k^{\mathrm{th}}$ variable.

We now state an important theorem regarding directional derivatives.

 Theorem 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ with $\mathbf{f} = (f_1, f_2, ..., f_m)$. Let $\mathbf{u} \in \mathbb{R}^n$. Then the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ in the direction of $\mathbf{u}$ exists if and only if the single variable vector-derivatives $f_k'(\mathbf{c}, \mathbf{u})$ exist for all $k \in \{ 1, 2, ..., m \}$ and $\mathbf{f}'(\mathbf{c} + h\mathbf{u}) = (f_1'(\mathbf{c}, \mathbf{u}), f_2'(\mathbf{c}, \mathbf{u}), ..., f_m'(\mathbf{c}, \mathbf{u}))$
• Proof: We write:
(1)
\begin{align} \quad \mathbf{f}'(\mathbf{c}, \mathbf{u}) &= \lim_{h \to 0} \frac{\mathbf{f}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c})}{h} \\ &= \lim_{h \to 0} \frac{(f_1(\mathbf{c} + h\mathbf{u}), f_2(\mathbf{c} + h\mathbf{u}), ..., f_m(\mathbf{c} + h\mathbf{u})) - (f_1(\mathbf{c}), f_2(\mathbf{c}), ..., f_m(\mathbf{c}))}{h} \\ &= \lim_{h \to 0} \frac{(f_1(\mathbf{c} + h\mathbf{u}) - f_1(\mathbf{c}), f_2(\mathbf{c} + h\mathbf{u}) - f_2(\mathbf{c}), ..., f_m(\mathbf{c} + h\mathbf{u}) - f_m(\mathbf{c}))}{h} \\ &= \left ( \lim_{h \to 0} \frac{f_1(\mathbf{c} + h\mathbf{u}) - f_1(\mathbf{c})}{h}, \lim_{h \to 0} \frac{f_2(\mathbf{c} + h\mathbf{u}) - f_2(\mathbf{c})}{h}, ..., \lim_{h \to 0} \frac{f_m(\mathbf{c} + h\mathbf{u}) - f_m(\mathbf{c})}{h} \right ) \\ &= ( f_1'(\mathbf{c}, \mathbf{u}), f_2'(\mathbf{c}, \mathbf{u}), ..., f_m'(\mathbf{c}, \mathbf{u})) \quad (*) \end{align}
• $\Rightarrow$ if $\mathbf{f}'(\mathbf{c} + h\mathbf{u})$ exists then each coordinate of the directional derivative exists and by $(*)$ this implies that the directional derivatives of the coordinate functions, $f_k'(\mathbf{c} + h\mathbf{u})$ exist for each $k \in \{1, 2, ..., m \}$.
• $\Leftarrow$ Suppose that $f_k'(\mathbf{c} +h\mathbf{u})$ exists for each $k \in \{1, 2, ..., m \}$. Then by $(*)$ each coordinate of $\mathbf{f}'(\mathbf{c} + h\mathbf{u})$ is finite and exists so $\mathbf{f}'(\mathbf{c} + h\mathbf{u})$ exists. $\blacksquare$
 Corollary 1: Let $S \subseteq \mathbb{R}^n$ be open, $\mathbf{c} \in S$, and $\mathbf{f} : S \to \mathbb{R}^m$ with $\mathbf{f} = (f_1, f_2, ..., f_m)$. Then the directional derivative of $\mathbf{f}$ at $\mathbf{c}$ with respect to the $k^{\mathrm{th}}$ variable exists if and only if the single variable vector-derivatives $D_k f_j(\mathbf{c})$ exist for all $j \in \{ 1, 2, ..., m \}$ and $D_k (\mathbf{f}(\mathbf{c}) = (D_k f_1(\mathbf{c}, D_k f_2(\mathbf{c}), ..., D_k f_m (\mathbf{c}))$
• Proof: Set $\mathbf{u} = \mathbf{e}_k$ where $\mathbf{e}_k = (0, 0, ..., 0, \underbrace{1}_{k^{\mathrm{th}} \: coordinate}, 0, ..., 0)$. Then by Theorem 1 we have that:
(2)
\begin{align} \quad \mathbf{f}'(\mathbf{c}, \mathbf{e}_k) & = (f_1'(\mathbf{c}, \mathbf{e}_k), f_2'(\mathbf{c}, \mathbf{e}_k), ..., f_m'(\mathbf{c}, \mathbf{e}_k)) \\ D_k \mathbf{f} (\mathbf{c}) & = (D_k f_1(\mathbf{c}), D_k f_2(\mathbf{c}), ..., D_k f_m (\mathbf{c}) \quad \blacksquare \end{align}